Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 21}{x - 5} = \dfrac{14x - 24}{x - 5}$
Solution: Multiply both sides by $x - 5$ $ \dfrac{x^2 + 21}{x - 5} (x - 5) = \dfrac{14x - 24}{x - 5} (x - 5)$ $ x^2 + 21 = 14x - 24$ Subtract $14x - 24$ from both sides: $ x^2 + 21 - (14x - 24) = 14x - 24 - (14x - 24)$ $ x^2 + 21 - 14x + 24 = 0$ $ x^2 + 45 - 14x = 0$ Factor the expression: $ (x - 9)(x - 5) = 0$ Therefore $x = 9$ or $x = 5$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.